 # Frank Morgan's Math Chat - MATHFEST '99: PRIMES AND PRIZES August 5, 1999

Last weekend over 1100 gathered in Providence for the summer MathFest of the Mathematical Association of America. Hedrick lecturer Carl Pomerance, recently of the University of Georgia and now at Bell Labs, with his marvelous understated humor, gave fascinating accounts of prime numbers. The largest known prime, discovered by Nayan Hajratwala on June 1, is 26,972,593 - 1, a "Mersenne" number of the form 2p - 1 (with the exponent p itself prime). It is an open question whether infinitely many Mersenne numbers are prime, or even whether infinitely many Mersenne numbers are not prime! It is well known, however, that there are infinitely many primes. Indeed, the Prime Number theorem says that the number of primes less than x is on the order of x/log x. The Riemann Hypothesis, perhaps the most important outstanding conjecture in mathematics, gives a more accurate estimate.

New prize winners included Ravi Vakil of MIT, who won the Trevor Evans award for his article on "The youngest tenured professor in Harvard's history" in Math Horizons (September, 1998). The article describes Noam Elkies's triumphs as a high school student in the Math Olympiad, his work in number theory, and his achievements in chess and music. More on the Math Olympiad appears in Vakil's book, A Mathematical Mosaic: Patterns and Problem-Solving.

The MAA Board of Governors discussed plans for Special Interest Groups in the MAA ("SIGMAAs") and for a new popular magazine.

US CHIEF JUSTICE WILLIAM REHNQUIST won a Washington Post quiz, "What auto is referred to in a license plate that reads 1 DIV 0?" with his answer:

"I believe it refers to an Infiniti, since when you divide 0 into 1, the result is infinity."

But is it legal to divide by 0? (Reuters, as reported in The Christian Science Monitor of 14 July 1999.)

OLD CHALLENGE (Arthur Pasternak). Abel and Beta are each given a box containing cash. It is known that one box contains 10 times the cash that is in the other box. They are give the opportunity to swap boxes. Abel reasons as follows:

Assume I have 10X in my box. Then the other box has either X or 100X with equal probability. If it has 100X, I will gain 90X by swapping; if it has X, I will lose 9X. Therefore I expect to gain 40.5X by swapping boxes.

This seems logical but, of course, Beta reasons the same way and concludes that she is better off swapping as well. Can they both be right?

ANSWER. No, by symmetry, each has expected gain 0 from swapping. The flaw in Abel's reasoning is that it is impossible for X and 10X to be equally likely! For example, if $1,$10, $100,$1000, . . ., all had the same positive probability, the total probability would be infinite instead of just 100%.

Elliot Kearsley and Joseph DeVincentis argue that if the possible values were say $1,$10, . . ., $1,000,000 and Abel and Beta had to agree to swap, they would never swap. Of course Abel wouldn't swap if he had$1,000,000 (so that Beta had $100,000); nor would Beta. Now if Abel had$100,000, he would know that Beta had either $1,000,000 (and Beta wouldn't swap) or$10,000; so Abel wouldn't agree to swap with $100,000; nor would Beta. Similarly if Abel had$10,000, he would know that Beta either had $100,000 (and wouldn't swap), or$1000; so Abel wouldn't agree to swap, and so on. Of course if Abel had just $1, he could agree to swap, but Beta with$10 would not, by the conclusion of the reasoning above.

(Actually the flaw in this argument is that Abel and Beta could have more complicated strategies of swapping with certain probabilities. Abel might decide to swap occasionally with $1,000,000 just to tempt Beta into swapping with$100,000 when Abel really had only \$10,000.)

This challenge partly resembles "The Wallet Paradox" described by Merryfield, Viet, and Watson in The American Mathematical Monthly of August-September 1997, in which Abel and Beta have to decide whether to swap wallets.

NEW CHALLENGE (via Charles Chace). Consider a statement of the form

(P and Q) => R if and only if (P => R) or (Q => R).

Is this a logical truth? What if

P is "ai is monotone"

Q is "ai is bounded"

R is "ai is convergent"

(where ai represents a sequence of real numbers).

Send answers, comments, and new questions by email to Frank.Morgan@williams.edu, to be eligible for Flatland and other book awards. Winning answers will appear in the next Math Chat. Math Chat appears on the first and third Thursdays of each month. Prof. Morgan's homepage is at www.williams.edu/Mathematics/fmorgan.

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